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@user-mf7vr3sj4u 45 минут бұрын
Can you explain why (at 1:05) the pressure forces act inwards as opposed to in the direction of flow / where in Chapter 2 specifically should I look to find out why?
@Tom-ts5qd Күн бұрын
Does this mean the Bernoulli Equation / Venturi effect can be applied to non ideal fluids as well? (approximately of course)
@FluidMatters 22 сағат бұрын
Yes. In real (viscous) fluids there are some energy losses. But "Bernoulli" is often still a good approximation for air and water, which both have low viscosity.
@francisfrancis4219 2 күн бұрын
A submerged ball when released does not “rocket out of the water”. It rockets up when in the water, but only to reach the surface.
@jacobhardy2313 8 күн бұрын
Great video and explanation
@FluidMatters 8 күн бұрын
Thanks. Glad to hear it was helpful.
@URKAEELIZACOLINPOWELL-ci1hw 15 күн бұрын
Is it the no slip condition that causes the velocity gradient, professor or the internal shear stresses?
@FluidMatters 13 күн бұрын
I'm not sure which part of this video you are specifically referring to. Here is a general answer. The fluid "sticks" to the surface i.e., the no slip condition. If the fluid above a stationary surface is moving (because of a pump or fan, for example), the fluid's viscosity will decrease (to zero) as you approach the surface. It's the internal viscous shear forces in the fluid, which opposes fluid motion, that slows the fluid in the near wall region. So, to answer your question, it is both effects. I hope that helps.
@URKAEELIZACOLINPOWELL-ci1hw 13 күн бұрын
So, hypothetically speaking, even if there doesn't exist 'no-slip' at the bottom surface ( but the fluid sticks to the top surface) there would be a velocity gradient in the fluid, solely because of the internal friction between them. Am I right, professor?
@URKAEELIZACOLINPOWELL-ci1hw 13 күн бұрын
I meant internal resistance between the fluid layers, alone, sir.
@user-pd5bn6qf5k 15 күн бұрын
when you have losses, you add them to get to EGL?
@FluidMatters 15 күн бұрын
Yes. The EGL remains constant only in the ideal friction case. But in the real world you always have energy losses in the flow direction. So, the EGL decreases because of energy losses (due to turbulence, wall viscous friction & pressure losses across a valve.) I hope that helps.
@ballerslounge2855 16 күн бұрын
Thanks for this lecture Dr Neylor i have a question though at 11:42 is the height difference not supposed to be (Z3 - Z2) or it doesnt matter
@FluidMatters 13 күн бұрын
Note that z is measured upward, i.e. Z2>Z3. You need that term to be positive, since you are moving downward into zone of higher pressure. So it must be Z2-Z3 to get a positive number. You might find it easier to work in terms of DeltaZ, where DeltaZ is always a positive value.
@john-hf9dt 18 күн бұрын
Thank you sir.
@antoniamillaraylizanaraban9380 20 күн бұрын
THIS VIDEO IS SO GOOD
@zhenccc 22 күн бұрын
All your videos helped a lot. Thank you so much!
@javierpicazo2107 25 күн бұрын
I like this series
@Kawsusstory 26 күн бұрын
Is that a plaquette on a hinge?
@Mehdiranjb 29 күн бұрын
🌹🌹
@TerminallyOnline92 Ай бұрын
At 13:29, shouldn’t it be C1 = -1/(2*mu*b)*dp/dx - U/b and not C1 = -b/(2*mu)*dp/dx - U/b ? Doing the algebra, I feel the b should be in the bottom of the fraction and not the top. Can someone explain please?
@FluidMatters Ай бұрын
Before simplification, we have b^2 in the numerator (from y^2 in the original expression applied at b). So, when you divide by b to isolate C1, you end up with b in the numerator. I hope that helps.
@TerminallyOnline92 Ай бұрын
@@FluidMattersThanks! I didn’t see that
@suleyman7478 Ай бұрын
great stuff
@RGCAD Ай бұрын
Great stuff, for the last problem i see we have our origin placed midway between the plates, I would therefore expect the y at the bottom to be y = -h
@FluidMatters Ай бұрын
Agree. The axis in the graphic (that I stole from the book publisher) is in the wrong place. But I think it's totally clear in the presentation that y=0 is at the bottom. At some point, a long time ago, I fixed it in the pdf download.
@baaa-ej7xj Ай бұрын
He saved my final exam😭😭😭
@comment8767 Ай бұрын
Point of Confusion for me - The cube as 6 faces with viscous stresses on each face. Therefore, there are 6 x 3 = 18 stresses on the differential element. Are we somehow saying that the stresses on opposite faces are the same, so that it is really only one? Then, all stresses are referred to a central point? What happens to the extra nine stresses?
@comment8767 Ай бұрын
OK, so you are not claiming that the nine stresses cover the entire differential volume. You are simply positing the stresses on three faces, and then using taylor series to get the stresses on the opposite face. The posited stress plus the Taylor-derived stress provide the stress on the entire cube.
@mohammadfallahzade2110 Ай бұрын
best instructor
@FluidMatters Ай бұрын
Thanks for the kind words. Best of luck with your studies.
@mohammadfallahzade2110 Ай бұрын
thank you you are a great teacher
@jamboreejackson8309 Ай бұрын
Hi professor, many thanks for the excellent lectures. I tried the derivation of the equation in cylindrical co-ordinates and I couldn't quite get there so ended up looking at another reference. Should the dA in the outlet term you show at 14:30 actually be (r+dr)dΘdz (rather than rdΘdz) to account for the slightly larger area at the outer radius? This is what the derivation I saw suggested.
@FluidMatters Ай бұрын
Good point. Yes, I think you are correct.
@mohammadfallahzade2110 Ай бұрын
thank you for your great explanation
@spiral-dz1eq Ай бұрын
from 3:40 on, why didn't we use the fluid for W and the wood for Fb?
@FluidMatters Ай бұрын
As shown in the free body diagram, W is the weight of the WOOD and Fb is the buoyancy force. Thus, Fb is the weight of the WATER displaced by the wood (Archimedes principle). That is why I used the specific gravity of wood for W, and the specific gravity of water for Fb. I hope that helps.
@spiral-dz1eq Ай бұрын
@@FluidMatters Thank you!
@andrewmerafuentes6683 Ай бұрын
I came just to check for a concept, then proceed to finish the whole series.
@FluidMatters Ай бұрын
Glad to be able to help. Best of luck with your studies.
@abdullahashraf8249 Ай бұрын
what da dog doing
@rashedshahariar2748 2 ай бұрын
Where is the vertical force acting on the top of the free body in no 2 prblm 7:55
@FluidMatters 2 ай бұрын
If I'm understanding your question: There is no vertical force on "the top of the free body". The vertical force acts upward and equals the weight of the water, which acts downward.
@rashedshahariar2748 2 ай бұрын
@@FluidMatters if we considered only the upper left quarter of the circle instead of the semi circle then there would have been vertical force acting on the upper part of the quarter circle then why not for the semi circle?
@rashedshahariar2748 2 ай бұрын
At 6:57 inclined forces at the upper portion should have vertical components ..isn't it? Can u pls clarify?
@gani82 2 ай бұрын
Continue the videos sir ..... following your videos ... ..waiting to learn from you sir.....
@FluidMatters 2 ай бұрын
Glad to hear you find them helpful. Good luck with your studies.
@colinwarn4606 2 ай бұрын
Just wanted to extend my thanks for putting together a great KZbin series on Fluid Mechanics. Been a couple years since I've taken the class, the series has been a great refresher on the theory, in addition to tying it to some very informative experiments that visually cement the concepts 🙂
@FluidMatters 2 ай бұрын
Thanks. Great to hear!
@anthonylorenzo1126 2 ай бұрын
Why isn't d^2u/dy^2 not 0?
@FluidMatters 2 ай бұрын
d2u/dy2=0 means that the velocity profile must be linear i.e. no curvature. There is no basis for this requirement, as you can see from the solution.
@jondory8134 2 ай бұрын
Nice job... Related to the FBD, do you need to include a Moment, M, along with the Fx & Fy? (assume a direction...?)
@FluidMatters 2 ай бұрын
This analysis is based on conservation of LINEAR momentum. So the are the forces needed to redirect the jet. No moment.
@KT-dv8qy 2 ай бұрын
THANKS A LOT SIR!!!!
@FluidMatters 2 ай бұрын
Glad to hear it was helpful. Best of luck with your studies.
@renatosureal 2 ай бұрын
To estimate the impact force of a 5-meter fall from a 90 kg person, we can use the work-energy principle which states that the work done by the impact force is equal to the change in kinetic energy. When a person falls from a height, they gain kinetic energy, which is then absorbed by the impact. The kinetic energy (KE) just before impact is given by: $$ KE = \frac{1}{2} m v^2 $$ Where: - \( m \) is the mass (90 kg in this case), - \( v \) is the velocity at impact. The velocity at impact can be found using the equation for velocity of an object in free fall: $$ v = \sqrt{2gh} $$ Where: - \( g \) is the acceleration due to gravity (approximately \( 9.81 \, m/s^2 \)), - \( h \) is the height of the fall (5 meters). Substituting the values, we get: $$ v = \sqrt{2 \times 9.81 \, m/s^2 \times 5 \, m} $$ $$ v = \sqrt{98.1 \, m^2/s^2} $$ $$ v \approx 9.9 \, m/s $$ Now we can calculate the kinetic energy: $$ KE = \frac{1}{2} \times 90 \, kg \times (9.9 \, m/s)^2 $$ $$ KE = \frac{1}{2} \times 90 \, kg \times 98.01 \, m^2/s^2 $$ $$ KE = 4410.45 \, J $$ The impact force (F) can then be estimated if we know the distance over which the impact force acts (d), which is the distance over which the person's momentum is brought to zero. This distance will depend on many factors, including how the person lands and the nature of the surface they land on. Assuming a certain stopping distance, we can use the formula: $$ F = \frac{KE}{d} $$ For example, if the person comes to a stop over a distance of 0.5 meters, the impact force would be: $$ F = \frac{4410.45 \, J}{0.5 \, m} $$ $$ F = 8820.9 \, N $$ So the impact force would be approximately **8820.9 Newtons**. Please note that this is a simplified calculation and the actual impact force can vary. For accurate results, especially for safety considerations or engineering applications, a detailed analysis considering all relevant factors would be necessary. Source: Conversation with Bing, 4/30/2024 (1) Impact Force Calculator - Calculate the impact force in a collision. www.gigacalculator.com/calculators/impact-force-calculator.php. (2) Impact Force - The Engineering ToolBox. www.engineeringtoolbox.com/impact-force-d_1780.html. (3) Free Fall Force Calculator Online. calculatorshub.net/physics-calculators/free-fall-force-calculator/. (4) Impact Force Calculator | Calculate Impact Force in Collision .... physicscalc.com/physics/impact-force-calculator/. (5) Impact Energy Calculator | Impact Force. www.omnicalculator.com/physics/impact-energy.
@FluidMatters 2 ай бұрын
ok
@ethanbaldwin1181 2 ай бұрын
Cool video!
@majroi 2 ай бұрын
Greetings from Turkey, this material helped me a lot. Thanks professor.
@ericjung7621 2 ай бұрын
hi professor, why are there no videos for content past chapter 5?
@FluidMatters 2 ай бұрын
That's where the intro fluid mechanics course ends at my university. I don't teach "Fluids II", at least for now.
@Frankchouzpy 2 ай бұрын
Impressive profound experiment and explanation. Additionally, I encountered numerous novel ideas in the comments section here.
@FluidMatters 2 ай бұрын
Thanks for the kind words. A colleague and I ended up writing a paper on this effect.
@Frankchouzpy 2 ай бұрын
These interesting videos make your tutorial more and more attractive. These formulas jump out of the screen and appear in our lives. Thank you for your amazing idea!
@Frankchouzpy 2 ай бұрын
Nice! Chapter 2 completed！！Next is the two ptional videos haha.
@Frankchouzpy 2 ай бұрын
I can’t stop learning, so take a break, click the like button, and then continue!
@mfn1311 2 ай бұрын
why do I have a course called wave physics and fluid mechanics? I think both are hard enough on their own 🥲
@FluidMatters 2 ай бұрын
Must be a different Dr. Naylor for the wave physics....
@salmankhanmohmand9174 2 ай бұрын
Sir where can i get its theory part...
@FluidMatters 2 ай бұрын
Here's the theory lecture for hydrostatic forces on curved surfaces: kzbin.info/www/bejne/gpXFdqOIjbqjhbMsi=DoySt3zv_nYTipHn You can find all the lectures as www.drdavidnaylor.net
@AtroXAir 2 ай бұрын
Great explanation! thank you very much
@Frankchouzpy 2 ай бұрын
Thanks for your great work on this playlist!!! I have learnt much from it. But there seems to be an integrate error at 21:45, the relationship of the p and p0 should include some exp. function or Ln function after the integration. Please check it. You have done such an excellent tutorial which is so useful for a student who is interested in the fluid mechanics! Thank you agian from China.
@FluidMatters 2 ай бұрын
The integration is correct. Notice there "z" in the dominator on the left side too (the adiabatic lapse rate). So, you get "ln" function on both sides. So, you end up taking the anti-log of both sides, after integration. The final results is a well-know equation -- you can just google Standard Atmosphere Equation, and you will find it is correct.
@Frankchouzpy 2 ай бұрын
@@FluidMatters Oh, I made a mistake. I didn't realize that the integrand on the left is the reciprocal of p instead of 1. I apologize for rushing my question, and thank you very much for your answer. Your tutorial is excellent, just like you. Bravo!
@batbayarbatsukh6393 2 ай бұрын
Greetings from Mongolia. Thanks for the clear explanation. Тhis helped me a lot I wish you good health.
@FluidMatters 2 ай бұрын
Mongolia! Hope to visit one day. Glad to hear the video was helpful.
@alisoltani7581 2 ай бұрын
Hello , what about the unsteady case? how can we sole this problem for u(t,y)?
@FluidMatters 2 ай бұрын
There are exact solutions for some simple unsteady problems, like an impulsively accelerated plane wall. This is beyond the undergrad level. See the classic book "Boundary-Layer Theory" by Schlichting, for example.
@alisoltani7581 2 ай бұрын
@@FluidMatters Ok,thank you very much🙏🙏
@Frankchouzpy 2 ай бұрын
Nice work thx. from China!😘
@schwarzehayvan8188 2 ай бұрын
Great Videos and Work. Thank you from Germany
@FluidMatters 2 ай бұрын
Glad you like them!
@masoudadli108 3 ай бұрын
the best of the best,, you make life easy..thanks from Sudan Africa
@FluidMatters 3 ай бұрын
Thanks for the kind words. Glad to hear the videos are helpful.
@lucasdovale9293 3 ай бұрын
Nice climb dude, those slabs look hard. Also, i was very happy to see both my hobbies in the channel, fluid mechanics and climbing!
@wheatlysparble7900 3 ай бұрын
thankyou so much